mcdonald (pam78654) HW 6B: Thermodynamics laude (89560) 1 This print-out should have 21 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points Calculate the change in enthalpy for the combustion of graphite using the data below. H 2 (g) + 1 2 O 2(g) H 2 O(l) H = 285.83 kj mol 1 CO 2 (g) + 2H 2 O(l) CH 4 (g) + 2O 2 (g) H = 882.00 kj mol 1 C graphite (s) + 2H 2 (g) CH 4 (g) H = 74.87 kj mol 1 C graphite (s) + O 2 (g) CO 2 (g) H =? 1. 385.21 kj mol 1 correct 2. 515.87 kj mol 1 3. 521.30 kj mol 1 4. 230.04 kj mol 1 5. 235.47 kj mol 1 In order for the three provided reactions to cancel to result in the unknown reaction (the combustion of graphite), the first reaction needs to be reversed and doubled, the second needs to be reversed, and the third will remain unchanged. Consequently, the overall change in enthalpy for the reaction is: H rxn = 2 285.83 + 1 882.00+ 74.87 = 385.21 kj mol 1 002 10.0 points Which of the reactions below will likely have the largest increase in entropy ( S rxn )? 1. C 5 H 12 (l)+8o 2 (g) 6H 2 O(g)+5CO 2 (g) correct 2. N 2 H 4 (g)+h 2 (g) 2NH 3 (g) 3. Na + (g)+cl (g) NaCl(s) 4. S 3 (g)+9f 2 (g) 3SF 6 (g) 5. 2CH 4 (g)+2o 3 (g) 4H 2 O(g)+2CO(g) The reaction with the greatest positive value for n gas will have the greatest value of S rxn. 003 10.0 points What is true about the first law of thermodynamics? 1. E univ < 0 2. E univ = 0 correct 3. E sys = 0 4. E sys < 0 5. E sys > 0 6. E univ > 0 Which of 004 10.0 points O 2 (g), O 2 (l), H 2 (g), H 2 (l), H 2 O(g), H 2 O(l) have a heat of formation equal to zero? 1. O 2 (g), O 2 (l), H 2 (g), H 2 (l), H 2 O(g), H 2 O(l) 2. O 2 (g), O 2 (l), H 2 (g), H 2 (l) 3. All of them, but only at absolute zero 4. O 2 (g), H 2 (g) correct
mcdonald (pam78654) HW 6B: Thermodynamics laude (89560) 2 5. O 2 (g), H 2 (g), H 2 O(g) Molecules in their native state at STP have a heat of formation of zero. 005 10.0 points If 25.0 g of water at 100.0 C are mixed with 15.0 g of water at 40.0 C, what temperature will the 40.0 g of combined water be at once they reach equilibrium? 1. 70.0 C 2. 77.5 C correct 3. 60.0 C 4. 62.5 C q cold = q hot 15 4.184 (T f 40) = 25 4.184 (T f 100) 15 (T f 40) = 25 (T f 100) 40 T f = 3100 T f = 77.5 C 006 10.0 points Calculate the reaction enthalpy for the formation 2Al(s)+3Cl 2 (g) 2AlCl 3 (s), of anhydrous aluminum chloride using the data 2Al(s)+6HCl(aq) 2AlCl 3 (aq)+3h 2 (g) H = 1049 kj HCl(g) HCl(aq) H = 74.8 kj H 2 (g)+cl 2 (g) 2HCl(g) H = 185 kj AlCl 3 (s) AlCl 3 (aq) H = 323 kj 1. 1883.5 kj 2. 1100.36 kj 3. 1225.7 kj 4. 1502.4 kj 5. 1450.85 kj 6. 1406.8 kj correct 7. 1826.2 kj Using Hess Law: The desired reaction is obtained by adding reaction1;6timesreaction2;3timesreaction 3; and 2 times the reverse of reaction 4: 2Al(s)+6HCl(aq) 2AlCl 3 (aq)+3h 2 (g) H = 1049 kj 6HCl(g) 6HCl(aq) H = 6( 74.8 kj) = 448.8 kj 3H 2 (g)+3cl 2 (g) 6HCl(g) H = 3( 185 kj) = 555 kj 2AlCl 3 (aq) 2AlCl 3 (s) H = 2(323 kj) = 646 kj 2Al(s)+3Cl 2 (g) 2AlCl 3 (s) H = 1406.8 kj 007 10.0 points The pressure-volume work done by an ideal gaseous system at constant volume is 1. P P 2. zero correct 3. E 4. q 5. V P When V = constant, nothing moves through a distance and therefore no work is done: w = 0. 008 10.0 points A CD player and its battery together do 500 kj of work, and the battery also releases 250
mcdonald (pam78654) HW 6B: Thermodynamics laude (89560) 3 kj of energy as heat and the CD player releases 50 kj as heat due to friction from spinning. Whatisthechangeininternalenergyof the system, with the system regarded as the battery and CD player together? 1. 700 kj 2. 200 kj 3. 800 kj correct 4. 750 kj 5. +200 kj Heat from the CD player is 50 kj. Heat from the battery is 500 kj. Work from both together on the surroundings is 250 kj. This question is testing your ability to see what the system is, and then look at ONLY the energy flow for the system. Here the systemisthebatteryandthecdplayertogether. U = q +w = [ 50 kj+( 250 kj)]+( 500 kj) = 800 kj 009 10.0 points Which of the following would probably have a positive S value? 1. He(g, 2 atm) He(g, 10 atm) 2. H 2 (g)+i 2 (s) 2HI(g) correct 3. 2NO 2 (g) N 2 O 4 (g) 4. O 2 (g) O 2 (aq) 5. 2Ag(s)+Br 2 (l) 2AgBr(s) We can predict the sign of S by noting the relative order of entropy: solids (lowest) < liquids < solutions < gases (highest) and the number of moles of each type. Also, if a gas is compressed (pressure increased) its entropy decreases. For the reactions given we have He(g, 2 atm) He(g, 10 atm) gas pressure increased S < 0 H 2 (g)+i 2 (s) 2 HI(g) 1 mol gas + 1 mol solid 2 mol gas; S > 0 2Ag(s)+Br 2 (l) 2AgBr(s) 2 mol solid + 1 mol liquid 2 mol solid; S < 0 O 2 (g) O 2 (aq) 1 mol gas 1 mol solution; S < 0 2NO 2 (g) N 2 O 4 (g) 2 mol gas 1 mol gas; S < 0 010 10.0 points Which of the following statements is always true? 1. An exothermic reaction is spontaneous. 2. If the number of moles of gas does not changeinachemicalreaction,then S 0 = 0. 3. A reaction for which delta S 0 is positive is spontaneous. 4. If H 0 and S 0 are both positive, G 0 will decrease as the temperature increases. correct H = (+) S = (+) G = H T S = (+) T (+) = (+) T As T increases, G decreases. 011 10.0 points For the reaction 2SO 3 (g) 2SO 2 (g)+o 2 (g) H r = +198 kj mol 1 and S r = 190 J K 1 mol 1 at 298 K. The forward reaction will be spontaneous at
mcdonald (pam78654) HW 6B: Thermodynamics laude (89560) 4 1. temperatures above 1042 K. correct 2. all temperatures. 3. temperatures above 1315 K. 4. temperatures below 1042 K. 5. no temperature. H r = +198 kj/mol S r = 0.019 G 0 = H 0 T S kj mol K G 0 < 0 for a spontaneous reaction, so 0 > H 0 T S T > H0 S 0 198 kj/mol = 0.019 mol K kj = 1042.11 K Thus the temperature would need to be > 1042.11 K. 012 10.0 points Which of the following is not a definition of enthalpy or change in enthalpy? 1. A measure of the motional energy of a system. correct 2. A correction for internal energy that accounts for pressure-volume work. 3. A measure of the heat of a system at constant pressure. 4. A measure of a system s ability to change the entropy of its surroundings. Total motional energy is a a measure of the internal energy (U, or sometimes E) of a system, not the enthalpy (H). 013 10.0 points Which of the following quantities is not path independent? 1. entropy (S) 2. volume (V) 3. pressure (P) 4. heat (q) correct 5. enthalpy (H) Heat quantitatively describes a transition between states, but not a state itself, and is thus a process quantity (path function), not a state function. 014 10.0 points Which of the following would experience the smallest increase in temperature if 1 kj of heat were added to it? 1. 1 g of copper metal 2. 10 g of water correct 3. 10 g of copper metal 4. 1 g of water Heat capacity is an extensive property, that reflects the amount of energy required to raise the temperature of an object. Because water has a much higher heat capacity than metals (to some extent a consequence of IMF), and because 10 grams is more than 1 gram, the 10 gram sample of water would experience a much smaller increase in temperature than the other choices. 015 10.0 points Given the following data: 3CO 2 (g)+4h 2 O(l) C 3 H 8 (g)+5o 2 (g) H = 1,110 kj mol 1 H 2 O(l) H 2 (g)+1/2o 2 (g) H = 142.5 kj mol 1 3C graphite +4H 2 (g) C 3 H 8 (g)
mcdonald (pam78654) HW 6B: Thermodynamics laude (89560) 5 H = 52 kj mol 1 calculate H for the reaction C graphite +O 2 (g) CO 2 (g) 1. 592 kj mol 1 2. 543 kj mol 1 3. 197 kj mol 1 correct 4. 543 kj mol 1 5. 1202 kj mol 1 H rxn = 11 3 (1,110 kj mol 1 ) 4 3 (142.5 kj mol 1 ) 1 3 ( 52 kj mol 1 ) = 197 kj mol 1 016 10.0 points Rank the following reactions from least to greatest in terms of change in entropy ( S rxn ): a) KNO 3 (aq)+nacl(aq) NaNO 3 (aq)+kcl(aq) b) 4Ag(s)+O 2 (g) 2Ag 2 O(s) c) 2H 2 O 2 (l) 2H 2 O(g)+O 2 (g) d) NaHCO 3 (s) NaOH(s)+CO 2 (g) 1. b < a < c < d 2. a < b < d < c 3. c < d < b < a 4. b < a < d < c correct 5. a < d < b < c 6. d < c < a < b Reaction b has a decrease in total moles and a decrease in moles of gas; S rxn is a negative number. In reaction a, no changes in phase and no changes in number of moles occur; S rxn will be nearly zero. Reaction d evolves a single mole of gas; S rxn will be a small positive number. Reaction c evolves 3 moles of gas; S rxn will be a large positive number. 017 10.0 points Phosphine (the common name for PH3, a highly toxic gas used for fumigation), has a H vap = 14.6 kj mol 1 and a S vap = 78.83 J mol 1 K 1. What is the normal boiling point of phosphine expressed in centigrade? 1. 185.2 C 2. 0.2 C 3. 87.8 C correct 4. 273 C Because boiling is an equilibrium process, G vap = 0 = H vap T S vap. And so T S vap = H vap and T = H vap / S vap = 14,600J mol 1 /78.83 J mol 1 K 1 = 185.2 K = 87.8 C 018 10.0 points Which of the following combustion reactions does not occur at high temperature? I) C 6 H 12 O 6 (s)+6o 2 (g) 6CO 2 (g)+6h 2 O(g) II) 2H 2 (g)+o 2 (g) 2H 2 O(g) III) C 3 H 8 (g)+5o 2 (g) 3CO 2 (g)+4h 2 O(g) 1. II, III 2. II only correct 3. III only
mcdonald (pam78654) HW 6B: Thermodynamics laude (89560) 6 4. I, II 5. I, II, III 6. I, III 7. I only Only reaction II results in a decrease in entropy, and is thus becomes less spontaneous and eventually non-spontaneous as temperature increases. 019 10.0 points (For this problem assume that the calorimeter itselfabsorbs no heat, thedensity of water is 1.00 g ml 1, and the specific heat capacity of water is 4.184 J g 1 K 1.) 1.14 g of octane (C 8 H 18 ) is combusted in a bomb calorimeter surrounded by 1 L of water. The initial and final temperatures of the water are 25 C and 38 C respectively. Determine the molar enthalpy of combustion of octane. 1. 5,440 kj mol 1 correct 2. 54,400 kj mol 1 3. 54.4 kj mol 1 4. 544 kj mol 1 5. 5.44 kj mol 1 T = T f T i = 38 C 25 C = 13 C = 12.98 K m = 1 L 1000 ml L n = 1.14 g octane 114 g mol 1.00 g ml = 1000 g H rxn = H cal = mc T mc T n = 0.01 mol 4.184 J 1000 g g K = 13 K 0.01 mol = 5,440 kj mol 1 020 10.0 points If both n gas and n for a particular reaction are zero, which statement below would be the most accurate prediction we could make about the change in entropy? 1. It is definitely positive. 2. It may be either positive or negative and is probably very small. correct 3. It is definitely negative. 4. It may be either positive or negative and is probably very large. Because the best way to predict the sign and magnitude of the change in entropy is based on the change in the moles of reactants and products (particularly of gaseous reactants or products), the sign would be impossible to predict, but the change would likely be very small. 021 10.0 points Which of the following statements concerning internal energy is/are true? I) If the expansion work is small, H and U are close in value. II) U for a system is equal to q at constant volume. III) Assuming no heat is exchanged, when pressure-volume work is done on the system, U is positive. 1. I, II, III correct 2. II only 3. III only 4. I, III 5. I only 6. II, III
mcdonald (pam78654) HW 6B: Thermodynamics laude (89560) 7 7. I, II StatementIfollowsfromtheidentity H = U + p V, because p V is the expansion work. Statement II follows from U = q+w, because w = 0 for processes that occur at constant volume. Statement III also follows from U = q +w, because w is the pressurevolume work. (Note: for reversible processes, expansion work and pressure-volume work are identical.)